HDU 3836 Equivalent Sets (tarjan求强联通分量)
2016-07-25 15:07
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题目链接:HDU 3836
题面:
Total Submission(s): 4074 Accepted Submission(s): 1433
[align=left]Problem Description[/align]
To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
[align=left]Input[/align]
The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
[align=left]Output[/align]
For each case, output a single integer: the minimum steps needed.
[align=left]Sample Input[/align]
4 0
3 2
1 2
1 3
[align=left]Sample Output[/align]
4
2
HintCase 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.
[align=left]Source[/align]
2011 Multi-University Training Contest 1 - Host by HNU
知识点:
这篇tarjan入门讲的很不错,可以用作入门。
题意:
给定一张有向图,问最少加几条边可以构成全图强联通,即强联通分量个数为1.
解题:
首先用tarjan算法进行染色缩点,强联通图中各个点的入度出度一定都大于0,若有点的入度为0,则肯定要加一条入边,若有点的出度为0,则肯定要加一条出边,分别统计出入度数为0的点的个数,取两者的大者作为结果。(肯定存在一种构造方式使得图可联通,还需特判原来就是一个联通块的情况,直接输出0即可)。
代码:
题面:
Equivalent Sets
Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 104857/104857 K (Java/Others)Total Submission(s): 4074 Accepted Submission(s): 1433
[align=left]Problem Description[/align]
To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
[align=left]Input[/align]
The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
[align=left]Output[/align]
For each case, output a single integer: the minimum steps needed.
[align=left]Sample Input[/align]
4 0
3 2
1 2
1 3
[align=left]Sample Output[/align]
4
2
HintCase 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.
[align=left]Source[/align]
2011 Multi-University Training Contest 1 - Host by HNU
知识点:
这篇tarjan入门讲的很不错,可以用作入门。
题意:
给定一张有向图,问最少加几条边可以构成全图强联通,即强联通分量个数为1.
解题:
首先用tarjan算法进行染色缩点,强联通图中各个点的入度出度一定都大于0,若有点的入度为0,则肯定要加一条入边,若有点的出度为0,则肯定要加一条出边,分别统计出入度数为0的点的个数,取两者的大者作为结果。(肯定存在一种构造方式使得图可联通,还需特判原来就是一个联通块的情况,直接输出0即可)。
代码:
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #define sz1 50010 #define sz2 20010 using namespace std; struct edge { int fm,to,nxt; }store[sz1]; int head[sz2],stack[sz2],dfn[sz2],low[sz2]; char vis[sz2]; int p,cnt,sig,tt,color[sz2],in[sz2],out[sz2]; void addedge(int fm,int to) { store[p].nxt=head[fm]; head[fm]=p; store[p].fm=fm; store[p++].to=to; } void Tarjan(int u) { vis[u]=1; low[u]=dfn[u]=cnt++; stack[++tt]=u; for(int i=head[u];~i;i=store[i].nxt) { int v=store[i].to; if(vis[v]==0)Tarjan(v); if(vis[v]==1)low[u]=min(low[u],low[v]); } if(dfn[u]==low[u]) { sig++; do { low[stack[tt]]=sig; color[stack[tt]]=sig; vis[stack[tt]]=-1; } while(stack[tt--]!=u); } } int main() { int n,m,a,b; while(~scanf("%d%d",&n,&m)) { p=cnt=tt=sig=0; memset(head,-1,sizeof(head)); memset(vis,0,sizeof(vis)); memset(out,0,sizeof(out)); memset(in,0,sizeof(in)); for(int i=0;i<m;i++) { scanf("%d%d",&a,&b); addedge(a,b); } for(int i=1;i<=n;i++) if(!vis[i]) Tarjan(i); if(sig==1) { printf("%d\n",0); continue; } for(int i=0;i<m;i++) { a=store[i].fm; b=store[i].to; if(color[a]!=color[b]) { out[color[a]]++; in[color[b]]++; } } a=b=0; for(int i=1;i<=sig;i++) { if(in[i]==0) a++; if(out[i]==0) b++; } printf("%d\n",max(a,b)); } return 0; }
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